How many cases of a three digit number $ABC$ exist such that the product of the two two-digit numbers $AB$ and $BC =$ the three digit number of $ABC$. $A, B,$ and $C$ represent the digits of the numbers in their respective places.
So far I've came up with the following equation:
$$100a + 10b + c = (10a + b)(10b + c),$$
but that would give you a messy equation with multiple variables that wouldn't really factor or simplify.
How else would you find out the number of possibilities of $ABC$ using only logic and reasoning?